Suppose that two identical objects go around circles of identical diameter, but one object goes around the circle twice as fast as the other. The turning (centripetal) force required to keep the faster object on the circular path is:
a) the same as the force required to keep the slower object on the path.
b) one-fourth as much as the force required to keep the slower object on the path.
c) half as much as the force required to keep the slower object on the path.
d) twice as much as the force required to keep the slower object on the path.
e) four times as much as the force required to keep the slower object on the path.
2007-09-15
17:06:04
·
2 answers
·
asked by
?
6
in
Science & Mathematics
➔ Physics
The answer is e. Think of the circle as a bent path with many sides. As the object runs around the path it must get little kicks which bend the path. Now if the object begins running around twice as fast you have to kick it twice as hard to bend the path by the same amount. So you might think the average force on the object would be doubled. But there is more. If it runs twice as fast it comes to each bend twice as often. So you have to kick it twice as hard twice as often. That increases the average force four times. If it goes around three times as fast you must kick it three times as hard three times as often. That would increase the average force nine times.
There is a road near my father's house with a curve in it. The highway sign warns 20 miles per hour, but one day my father decided to cheat a little and do 30 mph. What harm could an extra 10 mph do? In going from 20 to 30 he had only increased his speed 1.5 times. But increasing his speed 1.5 times increases the required...
2007-09-16
15:59:14 ·
update #1
centripetal force on his car 1.5 x 1.5 = 2.25 times. So a 50% speed increase requires a more than 100% increase in centripetal force -- which didn't happen on loose gravel and Dad's car ran off the road!
2007-09-16
16:00:37 ·
update #2